Reduced clique graphs of chordal graphs

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1 Reue lique grphs of horl grphs Mihel Hbib, Jurj Stho b,1,2,3 LIAFA-CNRS n Université Pris Dierot-Pris VII, Cse 7014, Pris Ceex 13, Frne b Wilfri Lurier University, Deprtment of Physis & Computer Siene, 75 University Ave W, Wterloo, ON N2L 3C5, Cn Abstrt We investigte the properties of horl grphs tht follow from the well-known ft tht horl grphs mit tree representtions. In prtiulr, we stuy the struture of reue lique grphs whih re grphs tht nonilly pture ll tree representtions of horl grphs. We propose novel eomposition of reue lique grphs bse on two opertions: ege ontrtion n removl of the eges of split. Bse on this eomposition, we hrterize steroil sets in horl grphs, n isuss horl grphs tht mit tree representtion with smll number of leves. Key wors: horl grph, lique tree, reue lique grph, spnning tree, lefge, steroil triple, steroil number, split-minor, forbien subgrph hrteriztion AMS lssifition: 05C15, 05C75 1. Introution A tree representtion of grph G onsists of host tree n olletion of its subtrees where the subtrees orrespon to the verties of G n two subtrees shre ommon vertex if n only if the orresponing verties re jent. A lique tree of G is tree representtion of G in whih the noes of the host tree orrespon to the mximl liques of G, i.e., for eh noe of the host tree, the set of verties of G whose orresponing subtrees ontin tht prtiulr noe is mximl lique of G. Equivlently, we n efine lique tree of G s tree whose verties re the mximl liques of G n who stisfies prtiulr onition (see Setion 1.1). It n be reily seen tht from every tree representtion of G, by (possibly) ontrting some eges of the host tree, one n lwys obtin lique tree of G. A grph is horl if it ontins no inue yle of length four or more. It is well-known tht horl grphs re preisely the grphs tht mit tree representtions [2, 6], or equivlently, the grphs tht mit lique trees. In this pper, we fous on this spet of horl grphs, n stuy properties of horl grphs ssoite with the struture of lique trees. In prtiulr, we investigte the reently reisovere notion of the reue lique grph of horl grph. This is the grph we obtin by tking the mximl liques of horl grph G s verties, n by putting eges between those verties for whih the orresponing liques interset in miniml seprtor tht seprtes them (see Setion 2 for preise efinition). This is subgrph of the (usul) lique grph ofg(where eges re between liques tht interset) n ws first efine in [5] more thn fifteen yers go. Sine then, this notion ws lrgely ignore (likely ue to the ft tht it ws nme lique grph The onferene version of this pper ppere s [10]. Emil resses: hbib@lif.jussieu.fr (Mihel Hbib), stho@s.toronto.eu (Jurj Stho) 1 Author initlly supporte by Fontion Sienes Mthémtiques e Pris. 2 Author further supporte by Kthie Cmeron n Chính Hoàng vi their respetive NSERC grnts. 3 Present ress: University of Wrwik, Mthemtis Institute, Zeemn Builing, Coventry, CV4 7AL, Unite Kingom Preprint submitte to Elsevier September 26, 2011

2 in [5] where the (more usul) lique grph ws lle lique intersetion grph ), n only reently [9, 19, 20], there hs been some interest in properties of this erive grph. The im of this pper is to further stimulte the interest in reue lique grphs by esribing some spets of their struture, explining where they iffer from similr (estblishe) notions, n in wht wy they n help us in solving omputtionl problems. Our ontribution onsists of two prts. Prt 1. Struture n eomposition of lbele reue lique grphs Sine t lest the ppers of Gvril [7] n Shibt [22], it hs been known tht the set of ll lique trees of horl grph G is preisely the set of ll mximum-weight spnning trees of the lique grph of G where the weight of n ege is efine s the size of the intersetion of the orresponing liques of G. As shown in [5] (lso see Theorem 3 in Setion 2), this is lso true for the reue lique grph of G but, unlike the (usul) lique grph, the reue lique grph of G hs the itionl property tht eh of its eges ppers in t lest one lique tree of G. (In other wors, the reue lique grph G is the (unique) union of ll lique trees of G.) At first glne, this might not seem like big vntge, but there re exmples [11, 19, 20] where it helps to el with the reue lique grph of G rther thn with its (usul) lique grph. To illustrte one of the funmentl ifferenes between the two lique grphs, note tht ing universl vertex to onnete horl grph G turns its lique grph into omplete grph wheres its reue lique grph remins unhnge. In the first prt of this pper (in Setion 2 n 3), we first esribe the generl struture of reue lique grphs bse on miniml seprtors. Then we propose prtiulr eomposition of reue lique grphs using the following two opertions: the ontrtion of n ege, n the removl of the eges of split. For these two opertions, we shll only onsier prtiulr eges n splits (s expline lter), n ll ny grph obtine using ombintion of the two opertions split-minor. It turns out, s we show, tht these opertions, when performe on the reue lique grph of horl grph G, orrespon iretly to opertions on G. In other wors, we show tht every split-minor of reue lique grph is gin reue lique grph. Moreover, we show tht in every reue lique grph there is t lest one ege n t lest one split tht we n ontrt or remove, respetively. This, in prtiulr, will imply these two opertions llow removing ll eges of every reue lique grph in some orer whih is lso true for eh of its split-minors (i.e., ny vli sequene of these two opertions n be extene to one tht removes ll eges). We feture this eomposition minly in the seon prt of this pper, but we envision tht it my fin lgorithmi uses in inutive onstrutions of solutions for problems on reue lique grphs from solutions on their split-minors. Prt 2. Use of reue lique grphs to hrterize properties of horl grphs In the seon prt of the pper (in Setion 4 n 5), we fous on prmeters of horl grphs, in prtiulr, those tht n be esribe in terms of their reue lique grphs. First, we fous on the notion of n steroil set n n steroil number. We sy tht set of verties A of grph G is steroil if for eh A, the verties in A \{} belong to ommon onnete omponent of G N[]; the steroil number of G, enote by (G), is then the size of lrgest steroil set of G. We strt by showing tht steroil sets (of verties) in horl grphs orrespon to steroil olletions of (mximl) liques (see Setion 4). This will llow to look t these sets purely from the perspetive of reue lique grphs n thus will lllow us to pply the notion of split-minor. In prtiulr, we show tht the steroil number of grph is monotone prmeter with respet to prtiulr (but rther tehnil) restrition of the notion of split-minor tht we ll goo splitminor (expline lter). Tht is, we show tht if the reue lique grph of G is goo split-minor of the reue lique grph of G, then (G) (G ). Note tht this will llow us to onlue tht there exists olletion of reue lique grphs suh tht horl grphgstisfies (G) < k if n only if the reue lique grph of G ontins no goo split-minor from the olletion. We then go on to prove tht suh olletion onsists preisely of one grph (up to the lbels of eges), nmely, the str with k eges. In the se k = 3, this will provie us with new hrteriztion of intervl grphs using single forbien obstrution. (Rell tht intervl grphs re the intersetion grphs of intervls of 2

3 the rel line, or equivlently, the horl grphs with (G) < 3 s shown in [16]). We remrk tht this is ifferent from other known hrteriztions of intervl grphs suh s the hrteriztion by forbien inue subgrphs whih ontins severl infinite fmilies (see Figure 6). Seonly, we look t nother prmeter of horl grphs G, the lefge l(g), whih is efine s the smllest number of leves in the host tree of tree representtion of G, or equivlently, the smllest number of leves in lique tree of G. We similrly pply the notion of split-minor to this onept. We show tht the lefge is lso monotone but for ifferent restrition of the notion of split-minor, n we further isuss smll substrutures tht use the lefge to be lrge. We remrk tht it is known tht both the steroil number n the lefge n be ompute in polynomil time for horl grphs by the respetive lgorithms from [13] n [11]. But it is the results presente in this pper tht provie us with struturl unerstning tht unerpins the two lgorithms. In prtiulr, the lgorithm in [11] is lrgely bse on the work from this pper, wheres the lgorithm from [13] n be, in ft, seen s trying to fin lbele k-str when eiing if the given grph hs steroil number t lest k. The interest in these two prmeters is minly ue to the ft tht if they re boune, we n effiienly solve severl problems on horl grphs tht woul otherwise be hr. For instne, the mximum inepenent ominting set [1] n brnhwith [21] n be solve effiiently if the steroil number, respetively, lefge is boune. Also, bnwith [14] n be effiiently 2k-pproximte on horl grphs of lefge t most k. We note tht these n similr results lrgely stem from the observtion tht the two prmeters n be seen s mesure of how fr horl grph is from being n intervl grph s the lefge n steroil number two orrespon preisely to intervl grphs. When one of the prmeters is boune, one n usully pt n effiient lgorithm tht works for intervl grphs to lso work on horl grphs where the prmeter is boune. Finlly, note tht, s this is struturl pper, we o not isuss omplexity issues here suh s the ones onnete to testing for the existene of (fixe) split-minor in given grph. Further, we o not nlyze the struture of unlbele reue lique grphs, tht is, the struture of grphs tht n be lbele to beome reue lique grphs. These issues re urrently subjet of n ongoing investigtion of the uthors n will be publishe seprtely. For nottion n stnr grph-theoretil notions use in this pper we refer the reer to [23] Clique tree A lique tree of onnete horl grph G is ny tree T whose verties re the mximl liques of G suh tht for every two mximl liques C,C, eh lique on the pth from C to C in T ontins C C. Further, the eges of T re lbele where ege CC of T hs lbel C C. (See Figure 1b) 1.2. Miniml seprtors A set S V(G) isonnets vertex from b in G if every pth of G between n b ontins vertex from S. A non-empty set S V(G) is miniml seprtor of G if there exist n b suh tht (i) S isonnets from b in G, n (ii) no proper subset of S isonnets from b in G. Observe tht the efinition implies tht the verties, b re neessrily in the sme onnete omponent of G, sine otherwise the empty set isonnets them n thus violtes (ii). For instne, the set S = {,} is miniml seprtor of the grph G in Figure 1, sine it isonnetsbnbut neither{} nor{} oes, beuse of the pths,,b n,,b, respetively. Similrly, S = {} is miniml seprtor, beuse it isonnets g n. This emphsizes tht, unlike wht its nme suggests, miniml seprtor is not neessrily n inlusion-wise miniml utset. Miniml seprtors ply n importnt role in horl grphs. In prtiulr, it is well-known tht every miniml seprtor of horl grph inues lique. Moreover, the onverse is lso true. Theorem 1. [8] G is horl if n only if every miniml seprtor of G is lique. 3

4 e ) f k g j h b i e b) f k g b j h b bi e f k g h b b bi j e f g b h b f g b h ) k j bi ) Figure 1: ) G, b) two exmple lique trees of G, ) C r(g), ) H {} 1.3. Seprting pirs As mentione bove, every miniml seprtor of horl grph G is lique, n hene, it is neessrily ontine in mximl lique of G. In ft, it turns out tht it is ontine in t lest two mximl liques, n moreover, it is preisely the intersetion of these liques. This is shown s follows (see lso Lemm 2.3 in [2]). Two mximl liques C,C of G form seprting pir if C C is non-empty, n every pth in G from vertex of C \C to vertex of C \C ontins vertex of C C. Theorem 2. A set S is miniml seprtor of horl grph G if n only if there exist mximl liques C,C of G forming seprting pir suh tht S = C C. Proof. If mximl liquesc,c ofgform seprting pir, thenc C isonnets vertex C\C from vertex of b C \ C. Moreover, C C is inlusion-wise miniml with this property, sine for every z C C, the verties,z,b form pth from to b. Finlly, C C is lso non-empty by the efinition of seprting pir, n hene, S = C C is miniml seprtor of G. Conversely, let S be miniml seprtor of G. Tht is, S is non-empty n there exist verties,b suh tht S isonnets from b in G n no proper subset of S hs this property. Let K n K b be the two onnete omponents of G S tht ontin n b, respetively. Let x be vertex of K with s mny neighbours in S s possible mong the verties of K. Suppose tht x is not jent to vertex w S. Sine S is miniml seprtor, w hs neighbour y in K. Let P be shortest pth from x to y in K. Sine w N(y) \ N(x), we hve on P onseutive verties x,y suh tht N(x) S = N(x ) S N(y ) S. In ft, by the mximlity of x, there exists w (N(x )\N(y )) S. We letp enote the subpth ofp fromx to y, n onlue thtd = w,p,w,w is yle in G where w,y re neighbours of x on this yle with w y E(G). So, horlity of G implies tht x must hve neighbour in D\{x,w,y }, whih is impossible beuse x is not jent to w n P is n inue pth. Therefore, we onlue tht N(x) S = S, n by the sme token, we hve vertexy in K b with N(y) S = S. So, if C n C re mximl liques of G tht ontin S {x} respetively S {y}, then C,C form seprting pir in G n S = C C s require. 4

5 2. The reue lique grph Immeitely from the respetive efinitions we hve tht every ege CC of lique tree of G is suh tht the liques C,C form seprting pir in G. This effetively inspires the following notion. The reue lique grph C r (G) of G is the grph whose verties re the mximl liques of G, n whose eges CC re between liques C,C forming seprting pirs. Further, s in the se of lique trees, the eges of C r (G) re lbele where ege CC of C r (G) hs lbel C C. (See Figure 1) We remrk tht if G is isonnete, it follows from the efinition tht the reue lique grph of G is the isjoint union of the reue lique grphs of the onnete omponents of G. It lso follows tht, onversely, if G is onnete, then its reue lique grph is lso onnete. Note tht, unlike wht is usul, we o not efine lique trees for isonnete grphs, n in the subsequent text, hving lique tree will lwys imply tht the onsiere grph n its reue lique grph re onnete. From this efinition, we n onlue tht if G is onnete, then every lique tree of G is spnning tree of the reue lique grph of G. Surprisingly, muh stronger sttement is true, whih ws lrey prove in [5], n it is the following funmentl result bout reue lique grphs. Theorem 3. [5] Let G be onnete horl grph. A tree T is lique tree of G if n only if T is mximum-weight spnning tree of C r (G) where the weight of eh ege CC is efine s C C. Moreover, the reue lique grph C r (G) is preisely the union of ll lique trees of G. Note tht by [4] it is known tht every onnete horl grph G hs t most V(G) mximl liques. Consequently, the number of noes in the reue lique grph of G n in every lique tree of G is t most V(G). In ontrst, the number of eges in the reue lique grph of G n be s lrge s Ω( E(G) 2 ); for instne, onsier G to be olletion of eges shring ommon vertex Seprtor grphs In orer to unerstn the struture of reue lique grphs, we now briefly stuy intersetions of mximl liques. As prove in Theorem 2, if two mximl liques re jent in the reue lique grph of G, then their intersetion is neessrily miniml seprtor (n hene it is non-empty). However, the onverse is not lwys true (see noes{f} n {b} in Figure 1) n, s we shll see, we nee to tke into ount prtiulr onnetivity onition. We utilize speil uxiliry grphs efine s follows. Let S V(G). The S-seprtor grph of G, enote by H S, is the grph whose verties re the mximl liques C of G with S C, n eges re between liques C,C suh tht C C S. In other wors, the -seprtor grph is preisely the lique (intersetion) grph of G, n if S, then H S is the intersetion grph of the mximl liques of G S whose verties re ompletely jent to S in G. For instne, if G is the grph from Figure 1, then the S-seprtor grph H S for S = {} is the grph epite in Figure 1. Note tht this grph hs three onnete omponents. Using the grphs H S, we now hrterize the struture of the reue lique grph C r (G) by proving tht two mximl liques re seprte by their intersetion S if n only if they re in ifferent onnete omponents of the S-seprtor grph H S. Theorem 4. Let G be horl grph, let C n C be two mximl liques of G, n let S = C C. ThenCC is n ege ofc r (G) with lbels if n only ifs is non-empty, nc nc belong to ifferent onnete omponents of H S. Proof. Suppose tht S is non-empty, n C n C re in ifferent onnete omponents of H S. Suppose tht CC is not n ege of C r (G). Then C,C is not seprting pir, n hene, sine S = C C, there exists pth x 1,...,x k from x 1 C \C to x k C \C suh tht x i S = C C for ehi {1...k}. It follows thtx,x 1,...,x k,x is yle in G for ehx S. ButGis horl whih implies tht eh vertex of S is jent to ll of x 1,...,x k. Now, let C 1 = C, C k+1 = C, n for eh 5

6 i {2...k}, let C i be mximl lique of G ontining {x i 1,x i } S. Clerly, the liques C 1,...,C k+1 ll belong to H S. Also, C i C i+1 is n ege of H S for eh i {1...k}, sine C i C i+1 S {x i } S. Hene, there is pth from C 1 = C to C k+1 = C in H S ontriting tht C,C re in ifferent onnete omponents of H S. Conversely, suppose tht CC is n ege of C r (G) with lbel S. Then C,C is seprting pir implying S. Suppose tht C,C belong to onnete omponent K of H S. Then there is pth C 1,...,C k in K with C 1 = C n C k = C. Sine C i C i+1 is n ege of H S, there exist vertex x i (C i C i+1 ) \ S for eh i {1...k 1}. Clerly, x 1 C n x k 1 C. Also, x i S = C C n x i x i+1 is n ege of G for eh i {1...k 2}. We onlue tht x 1,...,x k 1 is pth in G from vertex of C \C to vertex of C \C with no vertex of C C. But then CC is not n ege of C r (G), ontrition Struture of lique trees In the previous prgrphs, we hrterize the reue lique grph of G using miniml seprtors. Bse on this, we look bk t the reltionship between the lique trees n the reue lique grph ofg. By Theorem 3, every lique treet of onnete horl grphgis hrterize in terms of C r (G) s mximum-weight spnning tree of C r (G). Conversely, we hrterize C r (G) in terms of T s follows. Theorem 5. Let G be onnete horl grph, let T be lique tree of G, n let C,C be two mximl liques of G. Then CC is n ege of C r (G) with lbel S = C C if n only if there exists n ege with lbel S on the pth from C to C in T. Proof. Let C,C be two mximl liques of G n let S = C C. Let P enote the pth from C to C in T. First, suppose tht CC is n ege of C r (G) with lbel S. Then, by Theorem 4, the liques C n C belong to ifferent onnete omponents of H S. Moreover, sine T is lique tree, ll liques on P belong to H S. So, there must be onseutive liques C,C on P tht belong to ifferent onnete omponents of H S. By Theorem 4, the ege C C hs lbel S, n we re one. Conversely, suppose tht CC is not n ege of C r (G) but there is n ege C C with lbel S on P. We onlue tht S is miniml seprtor, n hene, Theorem 4 yiels tht C n C belong to the sme onnete omponent of H S. So, there exists pth P = C 1,...,C k from C 1 = C to C k = C in H S. By the efinition of P, we hve C i C i+1 S for eh i {1...k 1}. Now, we remove from T the egec C to obtin subtree T 1 tht ontins C n subtree T 2 tht ontinsc. Sine C 1 = C belongs to T 1 n C k = C belongs to T 2, there exists i suh tht C i belongs to T 1 while C i+1 belongs to T 2. Clerly, the pth from C i to C i+1 in T ontins the ege C C. So, sine T is lique tree, we onlue S = C C C i C i+1, ontrition. Therefore, no ege on P hs lbel S. Finlly, we n esribe the struture of T in terms of the grphs H S s follows. Theorem 6. Let G be onnete horl grph, n let T be lique tree of G. Let S be miniml seprtor of G, n let k S enote the number of onnete omponents of H S. Then T ontins extly (k S 1) eges with lbel S, n eh onnete omponent of H S inues onnete subgrph in T. Proof. Let C,C be two mximl liques of G tht belong to H S, n let P be the pth from C to C in T. We observe tht every lique on P lso belongs to H S, sine T is lique tree. It follows tht the verties of H S inue in T onnete subgrph. Next, suppose tht C,C belong to some onnete omponentk of H S. We show tht every lique on P lso belongs to K. Suppose otherwise, let C be the first lique on P tht lies outsie K. (Rell tht P is pth from C to C ). Let C be the lique on P just before C. Thus, C belongs to K, n hene, C C = S by Theorem 4. This implies tht C C = S beuse T is lique tree. Thus, by Theorem 5, CC is n ege of C r (G) with lbel S. But C,C both belong to K whih ontrits Theorem 4. Therefore, the verties of K inue onnete subgrph in T. 6

7 Finlly, sine both the verties ofh S n of eh onnete omponent ofh S inue subtree in T, it follows tht T must ontin extly (k S 1) eges with lbel S. Note tht if S is miniml seprtor of G, then k S 2. This n be seen s follows. By Theorem 2, there re mximl liquesc,c tht form seprting pir suh thtc C = S. Thus, CC is n ege of C r (G) with lbel S, n, by Theorem 4, the liques C n C re in ifferent onnete omponents of H S. This implies k S 2. So, s onsequene of the bove theorem, we hve tht every lique tree T of G ontins t lest one ege with lbel S whih lso shows tht G hs t most V(G) 1 miniml seprtors, sine T hs t most V(G) noes s mentione previously. 3. Split-minors In this setion, we fous on ifferent spet of the struture of reue lique grphs. We evelop eomposition tehnique tht will llow us to ompletely eompose ny reue lique grph using three nturl reution rules. We shll utilize this eomposition in lter setions. Let us strt by explining some intuition behin the eomposition. When onsiering lique trees (n tree representtions), it is nturl to perform ertin opertions on the (host) tree. In prtiulr, removing n ontrting eges of the tree re nturl opertions, sine they result in one or more lique trees tht represent speifi subgrphs or supergrphs of the originl grph. For this reson, the two opertions re sometimes use to onstrut ivie-n-onquer or greey lgorithms for problems on horl grphs. Sine we im to rgue tht reue lique grphs n be use lgorithmilly in ple of lique trees (s nonil substitute), we nee to fin similr opertions on reue lique grphs. Our gol therefore is the following: (G1) efine ege ontrtion on reue lique grphs, (G2) efine ege removl on reue lique grphs, n (G3) efine the two opertions so tht the orer in whih they re performe is inepenent of the result. First, let us look t ontrtion. Unlike the usul efinition, to formlly efine ontrtion in lique trees n reue lique grphs, we nee to be little reful. Note tht we efine lique trees n reue lique grphs s grphs whose vertex set is olletion of sets s well s the eges of these grphs re lbele with sets. Thus the ontrtion opertion on these grphs shoul reflet these fts. We efine it s follows. Let H be grph whose verties re sets n whose eges re lbele with sets. Let e = C 1 C 2 be n ege of H. Then we write H/ e for the grph obtine from H by ontrting e into C 1 C 2, tht is, (i) we remove the noes C 1 n C 2, new noe C 1 C 2, n (ii) for eh ege ofh with lbel S between some vertexc n one ofc 1, C 2, we new ege with lbel S between C n C 1 C 2. Note tht this opertion (s just eribe) my rete prllel eges. This is rel problem, n we shll el with it in moment. But before tht, let us look t lique trees first where prllel eges re not n issue with respet ege ontrtion (sine they o not ontin yles). LetT be lique tree of horl grphgn lete = C 1 C 2 be n ege oft. Observe tht thtt/ e is lso lique tree. Intuitively, we obtint/ e by treting the lique treet s tree representtion (i.e., host tree n its subtrees) n by ontrting e in the host tree n ll subtrees of the representtion tht ontin e. Clerly, this opertion oes not remove ny eges from G n only s new eges between the verties of C 1 n C 2. In prtiulr, T/ e is lique tree of the grph G we obtin from G by ing ll possible eges between the verties ofc 1 nc 2. This implies thtg is horl grph. the wor reution woul probbly be more pproprite here, but we woul like to voi the onfusion with the wor reue in reue lique grph. 7

8 b b b b f f e b e g g e b f? e e b b b f f b b b ef b ef b be b be b be b b e b e b b b b b eg bge f b f b f b f ) b) ) ) e) f) Figure 2: Exmples justifying the efinition of split-minor. b be b In reue lique grphs, s mentione bove, the opertion of ege ontrtion my rete prllel eges. Sine we woul like to el with simple grphs, we nee to eie how to remove prllel eges. Usully, one just removes ll but one ege between ny two verties to obtin n equivlent simple grph. Unfortuntely, prllel eges between sme verties my hve ifferent lbels. We nee to either ) eie whih of them to keep, or b) use rule to ssign lbel bse on the lbels of the prllel eges, or ) o not llow ontrtions tht proue prllel eges with ifferent lbels. Further, in nlogy to lique trees, we woul like tht the ontrtion of n ege C 1 C 2 in the reue lique grph of G proues the reue lique grph of the grph G obtine from G by ing ll possible eges between the verties of C 1 n C 2. It turns out tht ) is the only hoie tht lso stisfies this onstrint (s emostrte in Figure 2b). Thus this les us to the following efinition. We sy tht n egee = C 1 C 2 ofh is permissible if for every ommon neighbourc ofc 1 nc 2, the eges C 1 C n C 2 C hve the sme lbel. Clerly, if e is permissible, then ontrting it in H retes prllel eges only between C 1 C 2 n the ommon neighbours of C 1 n C 2. All these eges hve the sme lbel S, n we therefore remove ll but one of them to obtin simple grph. Consequently, we write H/ e for this (simple) grph from now on. In the sme fshion, we look t the seon opertion, the removl of eges. Agin, we rw inspirtion from lique trees. Let T be lique tree of G n let e = C 1 C 2 be n ege of T. Removing the ege e from T splits T into two onnete omponents; let X n Y be the vertex sets of those omponents (verties re mximl liques of G), n let V X be the union of liques C X, n V Y be the union of liques C Y. By gin treting T s tree representtion n restriting the host tree n subtrees to X, respetively Y, we observe tht T[X] n T[Y] re lique trees of G[V X ] n G[V Y ], respetively. Now, let us look t reue lique grphs. Unlike in lique trees, we nnot simply remove n ege in reue lique grph n expet the result to be gin reue lique grph or isjoint union thereof (see Figure 2). We hve to settle for the next best thing whih is removing eges of uts. (Note tht in trees removing eges n removing eges of uts re equivlent opertions.) A ut of grph H is prtition X Y of V(H) into two non-empty sets X n Y ; the eges of the ut X Y re the eges hving one enpoint in X n one enpoint in Y. A split of H is ut X Y of H suh tht every vertex of X with neighbour in Y hs the sme neighbourhoo in Y. (Note tht we llow X = 1 n Y = 1 unlike it is usul [3, 18] when efining split, sometimes lle 1-join). For ut X Y, we enote by V X the union of the sets in X n by V Y the union of the sets in Y. In n nlogy to lique trees, we only wnt to onsier those uts X Y of C r (G) for whihc r (G)[X] is the reue lique grph of G[V X ] n C r (G)[Y] is the reue lique grph of G[V Y ]. Moreover, to stisfy the onition (G3), we nee to mke sure tht n ege is permissible in C r (G)[X] or C r (G)[Y] if n only if it is permissible in C r (G). This ultimtely implies tht ll eges of the ut must hve the sme lbel, whih in turn yiels (by Theorem 4) tht the ut, in ft, must be split. For instne, 8

9 onsier the grph G in Figure 2; the prtition X = {{, b, },{b,, }}, Y = {{b, e},{, f}} epite in Figure 2e fils the first of the bove onitions, sine the reue lique grph ofg[v Y ] is notc r (G)[Y]. Further, onsier the the prtition X = {{,b,},{b,e},{,f}}, Y = {{b,,}} in Figure 2f; the ege {,b,}{,f} is permissible in C r (G)[X] but not in C r (G). This les to the following efinition. A split X Y of H is permissible if ll eges of the split hve the sme lbel. Clerly, if n ege e = C 1 C 2 in H[X] is permissible, then it is lso permissible in H, sine the eges between C 1, C 2 n their ommon neighbours in Y ll hve the sme lbel beuse X Y is permissible split. The sme hols for the eges in H[Y]. Now, ombining the bove opertions yiels the following notion. We sy tht grph H is split-minor of H if H n be obtine from H by (possibly empty) sequene of the following opertions: (L1) if v is n isolte vertex, remove v. (L2) if e is permissible ege, ontrt e. (L3) if X Y is permissible split, remove ll eges between X n Y. We remrk tht the rule (L1) is inlue to llow us to el with isonnete grphs. In the following prgrphs, we prove tht every split-minor of reue lique grph is gin reue lique grph. We then isuss permissible eges n splits in reue lique grphs, n s onsequene, we esribe how this implies eomposition of reue lique grphs. For the min theorem, we nee the following property of permissible eges. Lemm 7. Let G be onnete horl grph, let e = C 1 C 2 be permissible ege of C r (G), n let S = C 1 C 2 be the lbel of e. Then {C 1 } n {C 2 } re onnete omponents of H S. Proof. Clerly, both C 1 n C 2 re in H S beuse S = C 1 C 2. Suppose tht the omponent K tht ontins C 1 hs more thn one lique. By Theorem 6, the set K inues onnete subgrph in ny lique tree of G. Hene, C 1 hs neighbour C in K suh tht C 1 C is n ege of C r (G). This implies C 1 C S by Theorem 4, beuse C 1,C belong to the sme omponent of H S. Also, C 1 n C 2 re in ifferent omponents of H S beuse C 1 C 2 is n ege of C r (G) with lbel S. Hene, C n C 2 re in ifferent omponents of H S, n we onlue, by Theorem 4, tht C C 2 is lso n ege of C r (G) with lbel S. This retes tringle C 1,C 2,C in C r (G). However, C 1 C n C 2 C hve ifferent lbels, n hene, e = C 1 C 2 is not permissible ege, ontrition. We therefore onlue tht K = {C 1 } is onnete omponent of H S. By the sme token, {C 2 } is onnete omponent of H S. We re lmost rey to proof the min theorem of this setion. It only remins to isuss prtiulr tehnil subtlety of the sttement. We woul like to prove tht if we remove eges of permissible split X Y of the reue lique grph of G, we obtin nother reue lique grph, nmely, the reue lique grph of G = the isjoint union of G[V X ] n G[V Y ] (where V X n V Y re efine s before). The problem is tht some verties of G my belong to both V X n V Y, n s suh, they will pper in liques of both G[V X ] n G[V Y ], n pper both on eges of C r (G)[X] n C r (G)[Y]. However, in G eh vertex of V X V Y is represente by two istint verties. To fix this, we ugment the opertion (L3) to o the following fter removing the eges between X n Y. For every b V X V Y, we reple b by b in eh lique C Y tht ontins b n upte the lbels of the ffete eges (so tht the lbel of eh ege inites the intersetion of the two liques tht re the enpoints of the ege). Then when onsiering G, the isjoint union of G[V X ] n G[V Y ], we impliitly ssume tht the verties of V X V Y re reple in G[V Y ] by their prime ( ) opies. This llows us to sfely use the opertion (L3). With this in min, we n finlly ive into the proof of the theorem. the nme ws hosen in nlogy with the notion of minor where we remove, ontrt eges, n remove isolte verties. 9

10 Theorem 8. IfH is split-minor of C r (G), then there exists horl grphg suh tht H = C r (G ). Proof. By inution. Let H be the grph obtine from C r (G) by one of the three opertions. Cse 1. We pply the rule (L1) to n isolte vertex of C r (G). Then this vertex orrespons to mximl lique C of G tht forms onnete omponent of G, n hene, C r (G C) = H. For the rules (L2) n (L3), we observe tht if C r (G) is isonnete then the opertions (L2) n (L3) only ffet one of its onnete omponent while the other omponents remin the sme. In ft, n ege is permissible in C r (G) if n only if it is permissible in some onnete omponent of C r (G), n for every split of C r (G), there exists n equivlent split of onnete omponent of C r (G) where, in prtiulr, the eges of two splits re the sme. Also, by efinition, C r (G) is the isjoint union of the reue lique grphs of the onnete omponents of G, n it is onnete if n only if G is. This implies tht it suffies to prove the remining ses for onnete grphs. Thus, in wht follows, we shll ssume tht both G n C r (G) re onnete. Cse 2. We pply the rule (L2) to permissible ege e = C 1 C 2 of C r (G) with lbel S = C 1 C 2. In other wors, H = C r (G)/ e. Let G be the grph we obtin from G by ing ll possible eges between the verties of C 1 n C 2. We show G is horl n C r (G ) = C r (G)/ e. First, we look t the vertex sets of the two grphs. SineGis onnete, there exists, by Theorem 3, lique treet ofgtht ontins the egee. By tretingt s tree representtion n by ontrting e is its host tree n ll its subtrees, we onlue tht T/ e is lique tree of G. This implies tht G is horl, n tht the vertex set of T/ e is the sme s the vertex set of C r (G ). Also, by efinition, the vertex sets of T/ e n C r (G)/ e re the sme, sine the vertex sets of T n C r (G) re the sme (T is lique tree of G). This proves tht the vertex sets of C r (G ) n C r (G)/ e re the sme. It remins to onsier eges. (1) Every ege of C r (G ) is lso n ege of C r (G)/ e. Let CC be n ege of C r (G ), n suppose tht CC is not n ege of C r (G)/e. If neither of C,C is the lique C 1 C 2, then CC is not n ege of C r (G), n hene, there is pth P in G from vertex of C \ C to vertex of C \ C with no vertex of C C. However, G is subgrph of G, n hene, P is pth in G implying tht CC is not n ege of C r (G ), ontrition. So we my ssume C = C 1 C 2. We onlue tht we hve no C (C 1 \C 2 ) or no b C (C 2 \C 1 ), sine otherwise,b is pth from vertex of C 1 \C 2 to vertex of C 2 \C 1 implying tht C 1 C 2 is not n ege of C r (G), ontrition. So, without loss of generlity, we my ssume C C 2 C 1. Sine CC is not n ege of C r (G)/ e, both CC 1 n CC 2 re not eges of C r (G), n hene, there exists in G pth P from vertex of C \C 1 = C \(C 1 C 2 ) to vertex of C 1 \C with no vertex from C C 1 = C (C 1 C 2 ). But then P is lso pth in G implying tht CC is not n ege of C r (G ), ontrition. (2) Every ege of C r (G)/ e is lso n ege of C r (G ). Let CC be n ege of C r (G)/ e, n suppose tht CC is not n ege of C r (G ). This implies tht there exists pth in G between vertex of C \C n vertex of C \C with no vertex in C C. Let P be shortest suh pth, n let C \C n b C \C be the enpoints of P. First, suppose tht C is the lique C 1 C 2. The minimlity of P implies tht b is the only vertex of C 1 C 2 on P. Without loss of generlity, suppose tht b C 1. This implies tht CC 1 is not n ege of C r (G) beuse of the pth P. Hene, Lemm 7 implies tht C is not in H S, sine otherwise C n C 1 re in ifferent omponents of H S ontriting Theorem 4. Therefore, there exists S \C, n we onlue tht CC 2 is not n ege of C r (G) beuse P, is pth from C \C 2 to C 2 \C with no vertex of C C 2. But then CC is not n ege of C r (G)/ e, ontrition. So we my ssume tht neither ofc,c is the liquec 1 C 2, n therefore,cc is n ege ofc r (G). It follows tht some ege xy of P oes not belong to G. We must onlue x,y C 1 C 2, n without loss of generlity, we ssume x C 1 \ C 2, y C 2 \ C 1, n the verties,x,y,b pper on P in this orer. We further onlue tht no vertex of C 1 C 2 other thn x,y belongs to P, beuse P is n inue pth in G. In prtiulr, P ontins no vertex ofs n ontins extly two verties ofc C. 10

11 Now, let P 1 n P 2 enote the subpths of P from to x n from y to b, respetively. If there exists S \(C C ), then P 1,,P 2 is pth of G ontining no vertex of C C implying tht CC is not n ege of C r (G), ontrition. Hene, we must onlue S C C, n so, C belongs to the grph H S. In prtiulr, C n C 1 belong to ifferent onnete omponents of H S by Lemm 7, n hene, C 1 C is n ege of C r (G) with lbel S by Theorem 4. Tht is, C 1 C = S whih implies tht P 1 ontins no vertex of C 1 C. (Rell tht P ontins no vertex of S.) However, then P 1 is pth of G between C \C 1 n x C 1 \C, n hene, CC 1 nnot be n ege of C r (G), ontrition. Thus ombining (1)-(2) yiels C r (G ) = C r (G)/ e = H s require. Cse 3. We pply the rule (L3) to permissible split X Y of C r (G) where S is the (ommon) lbel of the eges between X n Y. Let V X V(G) enote the union of mximl liques in the set X, n let V Y V(G) enote the union of mximl liques in the set Y. Let G be the isjoint union of G[V X ] n G[V Y ]. We show tht C r (G ) = H. Tht is, we show C r (G[V X ]) = C r (G)[X] n C r (G[V Y ]) = C r (G)[Y]. First, we isuss the sets V X,V Y. In prtiulr, we show the following property. (3) V X V Y = S n there re no eges in G between the verties in V X \S n the verties in V Y \S. Sine every vertex of G belongs to t lest one mximl lique of G, we observe tht every vertex of G belongs either to V X or V Y or both. Further, sine there is t lest one ege CC in C r (G) where C X n C Y n this ege hs lbel S, we hve S C V X n S C V Y. Thus S V X V Y. To prove tht S V X V Y, suppose for ontrition tht there exists x (V X V Y ) \ S. Sine x V X V Y, there is lique C X with x C, n lique C Y with x C. LetT be lique tree of G, n let P be the pth of T between C n C. Sine C X n C Y, there exist onseutive liques C,C on P with C X n C Y. So, x C C, beuse x C C n T is lique tree. Further, T is subgrph of C r (G) by Theorem 3, n X Y is permissible split. Hene, C C is n ege of C r (G) with lbel S. But then x S = C C, ontrition. Finlly, we show tht there re no eges between V X \ S n V Y \ S. If otherwise, there re jent verties x V X \S n y V Y \S, n so there exists mximl lique C of G with x,y C. If C X, then y C V X, n hene, y V X V Y = S, ontrition. So, C Y but then x V X C V X V Y = S, ontrition. Now, we re rey to prove C r (G ) = H. By symmetry, it suffies to show C r (G[V X ]) = C r (G)[X]. (4) X is the set of mximl lique of G[V X ]. First, onsier lique C X. Sine C is lique of G[V X ] by the onstrution of V X, it is lso mximl lique of G[V X ], beuse it is mximl lique of G. Conversely, let C be mximl lique of G[V X ]. Clerly,C is lique ofg, n hene, there exists mximl lique C of G with C C. Rell tht we hve t lest one ege in C r (G) betweenx n Y. Hene, t lest one lique of G[V X ] properly ontinss, n so C\S, beuse C is mximl lique ofg[v X ]. Also, rell tht V X V Y = S. So, C X, sine otherwise we hve C V Y, n hene, C = C C V X V Y = S, ontrition. But then C V X, n hene, C = C beuse C is mximl lique of G[V X ]. So, we onlue C X. This shows tht C r (G)[X] n C r (G[V X ]) hve the sme vertex set. It remins to onsier eges. (5) CC is n ege of C r (G)[X] if n only if it is n ege of C r (G[V X ]). First, let CC be ny ege of C r (G)[X]. Then CC is lso n ege of C r (G[V X ]), sine ny pth in G[V X ] between vertex of C \C n vertex of C \C is lso pth in G. Conversely, let CC be n ege of C r (G[V X ]). Suppose tht CC is not n ege in C r (G)[X]. Then there is pth in G between vertex C \ C n vertex b C \ C tht ontins no vertex of C C. Let P be shortest suh pth. Sine CC is n ege of C r (G[V X ]), t lest one vertex of P lies outsie V X. Let x n y (possibly x = y) be respetively the first n the lst vertex on P outsie V X. Sine,b V X, we hve on P vertex x just before x, n vertex y right fter y. By the hoie of x,y, we onlue x,y V X. Also, x,y V Y \ S sine x,y V X n S V X. So, sine there re no eges betweenv X \S n V Y \S, we must onluex,y S. But then P is not n inue pth, sine S is lique by Theorem 1, ontrition. 11

12 So, ombining (4) n (5) yiels C r (V X ) = C r (G)[X]. By symmetry, lso C r (V Y ) = C r (G)[Y], n hene, we obtin C r (G ) = H s lime. Tht onlues the proof. We now show tht every reue lique grph ontins permissible eges n splits. We sy tht n egeeof C r (G) is mximl if there is no egee in C r (G) whose lbel stritly ontins the lbel of e. Similrly, n ege e is miniml if there is no ege e whose lbel is stritly ontine in the lbel of e. We note in pssing tht these two types of eges orrespon respetively to the mx-min n minmin seprtors of [9] whih ply prtiulr role in the forbien inue subgrph hrteriztion of pth grphs in [17]. Theorem 9. Every mximl ege e in C r (G) is permissible, n for every miniml ege e in C r (G), there exists permissible split X Y of C r (G) suh tht e is n ege between the sets X n Y. Proof. Firstly, note tht if n ege e is mximl, miniml, or permissible in C r (G), then it is lso respetively mximl, miniml, or permissible in some onnete omponent of C r (G). Further, permissible split of the onnete omponent of C r (G) ontining e n be ugmente to permissible split of C r (G) by rbitrrily ing other onnete omponents of C r (G) to one or the other sie of the prtition. This implies tht it suffies to prove the theorem for onnete grphs G. Lete = C 1 C 2 be mximl ege ofc r (G), n letc 1,C 2,C be tringle inc r (G) suh tht the eges C 1 C n C 2 C hve ifferent lbels. Tht is, if we enote S = C 1 C 2, S 1 = C C 1, n S 2 = C C 2, then we hve S 1 S 2. If S 1 S n S 2 S, then we onlue S 1 = S 2 = C C 1 C 2. So, without loss of generlity, we ssume tht there is S 1 \S. If lso b S\S 1, then,b is pth between vertex of C \C 2 n vertex of C 2 \C with no vertex of S 2 = C 2 C, but then C 2 C is not n ege of C r (G). We onlue S S 1. However, e is mximl ege, n hene, S 1 \S =, ontrition. Next, let e = C 1 C 2 be miniml ege of C r (G) with lbel S = C 1 C 2. Sine G is onnete, there exists, by Theorem 3, lique tree of T tht ontins the ege e. If we remove the ege e from T, we obtin two subtrees; let X n Y enote the vertex sets of these two subtrees suh tht C 1 X n C 2 Y. Further, letx 0 X enote the set ll liques C X with S C, n let Y 0 Y be the liques C Y with S C. We show tht if C X n C Y, then CC is n ege of C r (G) with lbel S if n only if C X 0 n C Y 0. This will prove tht X Y is permissible split of C r (G) s require. Consier C X n C Y. Sine T is lique tree, we hve C C C 1 C 2 = S. Hene, if CC is n ege of C r (G), we must onlue C C = S by the minimlity of e, n therefore, C X 0 n C Y 0. Conversely, if C X 0 n C Y 0, then C C = S beuse C S, C S, n C C S. Moreover, the ege e lies on the pth in T from C to C. Hene, by Theorem 5, we onlue tht CC is n ege of C r (G). We remrk tht not every permissible ege of C r (G) is neessrily mximl; for instne, the ege in Figure 1 lbele {} between the liques {,g} n {,h} is permissible, but it is not mximl beuse {,} is the lbel of the ege between the liques {,,} n {,b,}. We sy tht grph is totlly eomposble by set of rules if there is sequene of pplitions of the rules tht reues the grph to the empty grph (the grph with no verties). Combining the previous two results, we n now onlue the following. Theorem 10 (Split-minor eomposition). Every reue lique grph is totlly eomposble by the rules (L1) n (L2), n is lso totlly eomposble by the rules (L1) n (L3). Proof. Let H = C r (G) be miniml ounterexmple to the lim. Clerly, H hs t lest one vertex. IfH ontins no eges, then it hs n isolte vertex, n we n pply (L1). So, H hs n ege, whih implies tht it lso hs miniml ege e n mximl ege e. But then Theorem 9 yiels tht e is permissible, n hene, we n pply (L2). Also, H ontins permissible split X Y where e is one of the eges between X n Y, n we n pply (L3). So, H is not miniml ounterexmple. 12

13 We lose this setion by noting n interesting onnetion between split-minors of reue lique grphs n lgorithms for fining mximum-weight spnning tree. If G is onnete, we n, by the bove theorem, itertively ontrt mximl eges in C r (G) until the grph reues to single noe. Clerly, the ontrte eges yiel spnning tree T of C r (G), n it is not iffiult to see tht T is, in ft, lique tree of G. So, if t eh step we hoose n ege tht is not only mximl but lso hs lrgest lbel, we onstrut the sme tree tht Kruskl s lgorithm [15] woul onc r (G). In other wors, we onstrut mximum-weight spnning tree. Conversely, if T is lique tree of G, we n itertively ontrt in C r (G) the eges of T in eresing orer of their size, n eh suh ege is permissible t the time when we ontrt it. This provies us with n lterntive n lgorithmi proof of Theorem 3. A similr onlusion n be rehe by itertively removing permissible splits bse on eges with smllest lbels. If X Y is permissible split of C r (G), then lique tree of G n be obtine by fining lique trees for the horl grphs orresponing to C r (G)[X] n C r (G)[Y], n then ing ny ege from the eges of C r (G) between X n Y. This behviour n be observe in the Reverseelete lgorithm [12] for fining mximum spnning tree, in whih we remove minimum-weight eges unless they isonnet the grph; in our se, we remove ll but one ege between X n Y. 4. Asteroil number In this setion, we use the eomposition presente in Setion 3, to hrterize the steroil numbers of horl grphs. Rell tht set A of verties of G is steroil if for ny A, ll verties of A\{} belong to one omponent ofg N[], n the steroil number(g) of G is the size of lrgest steroil set in G. Note tht if G is isonnete, then it follows from the efinition tht the steroil number of G is the mximum over the steroil numbers of its onnete omponents Asteroil olletion of liques First, we show tht the steroil number of horl grph epens solely on the struture of its reue lique grph. We o this by relting it to similr notion efine on reue lique grphs. We sy tht n ege e of C r (G) is hit by lique C if some ege of C r (G) inient to C hs the sme lbel s e. We sy tht pth P of C r (G) is hit by lique C if some ege on P is hit by C. Otherwise, we sy tht P is misse by C, or tht C misses P. For instne, in Figure 1, the pth P onsisting of liques{,j},{,,},{,,f} is hit by the lique {b,,i}, beuse the lbel of the ege between {,j} n {,,} is the sme s the lbel of the ege between {,j} n {b,,i}. However, P is misse by the lique {,g}, sine no ege of P hs lbel {}. A olletion A of liques of G is n steroil olletion of liques if for ll istint C,C,C A, there is pth in C r (G) from C to C misse by C. As the reer woul now expet, we shll prove, in wht follows, tht G ontins n steroil set of size k if n only if C r (G) ontins n steroil olletion of liques onsisting of k liques. Before tht, we nee to show ouple of useful properties. First, we remrk tht we shll impliitly mke use of the following observtion. Lemm 11. An ege e of C r (G) with lbel S is hit by lique C if n only if S C. Proof. If e is hit byc, then some ege inient toc hs lbel S implyings C. Conversely, ifs C, thenc belongs to H S. Sine e hs lbel S, there re, by Theorem 4, t lest two onnete omponents in H S. Hene, if C is ny lique in onnete omponent of H S ifferent from tht of C, then, by Theorem 4, CC is n ege of C r (G) with lbel S. So, e is hit by C, sine CC hs the sme lbel s e. Next, using the following lemm will llow us to fous on lique trees only. Lemm 12. LetGbe onnete horl grph, lett be lique tree ofg, n letc 1,C 2,C 3 be istint mximl liques of G. Then every pth between C 1 n C 2 in C r (G) is hit by C 3 if n only if the pth in T between C 1 n C 2 is hit by C 3. 13

14 Proof. The forwr iretion is obvious, sine T is spnning tree of C r (G) by Theorem 3. For the bkwr iretion, let P be pth in C r (G) between C 1 n C 2 misse by C 3, n let e be n ege on the pth of T between C 1 n C 2 tht is hit by C 3. Let S enote the lbel of e. We onlue S C 3. Now, lett 1,T 2 enote the two subtrees oft we obtin by removing the egeefrom T wherec i belongs to T i for i = 1,2. Sine C 1 is in T 1 but C 2 is in T 2, there exists on P n ege e hving one enpoint in T 1 n the other in T 2. Let S enote the lbel of e. We observe tht the pth of T between the enpoints of e ontins the ege e, sine the enpoints re not both in T 1 or both in T 2. It follows tht S S C 3, beuse T is lique tree. So, C 3 hits e, but then C 3 lso hits P by Lemm 11. Finlly, we re rey to prove the theorem vertise erlier. Theorem 13. (G) k if n only if C r (G) ontins n steroil olletion A of liques with A = k. Proof. Firstly, if G is isonnete, then lso C r (G) is isonnete, sine C r (G) is the isjoint union of the reue lique grphs of the onnete omponents of G. In prtiulr, ny steroil olletion of liques in C r (G) is, by efinition, n steroil olletion in some onnete omponent of C r (G). Further, rell the steroil number of G is the mximum over steroil numbers of the onnete omponents of G. This implies tht it suffies to prove the theorem for onnete grphs G. Suppose tht (G) k, n let A = { 1,..., k } be n steroil set of G of size k. For eh i {1...k}, let C i enote ny mximl lique of G tht ontins i. Clerly, i C j n C i C j for eh i j, sine no two verties of A re jent. We show tht A = {C 1,...,C k } stisfies the lim. Suppose otherwise, n without loss of generlity, ssume tht every pth from C 1 to C 2 in C r (G) is hit by C 3. In prtiulr, if T is lique tree of G, then the pth of T between C 1 n C 2 ontins n ege e hit by C 3. Let S enote the lbel of e. We onlue S C 3. Now, let T 1,T 2 enote the two subtrees of T we obtin by removing the ege e from T where C i belongs to T i for i = 1,2. Rell tht A is n steroil set. So, there exists pth P in G from 1 to 2 suh tht 3 hs no neighbour on this pth. Sine 1 belongs to lique in T 1 n 2 belongs to lique in T 2, let us enote by y the first vertex on P tht belongs to lique in T 2. Note tht y 1, sine otherwise 1 belongs to both lique int 1 n in T 2 implying 1 S C 3 beuset is lique tree, ontrition. Consequently, we hve vertex x before y on P, n x is in no lique of T 2 by the minimlity of y. Sine x,y re onseutive on P, we hve xy E(G), n hene, there is lique in T 1 tht ontins both x,y. So y S C 3, beuse y lso belongs to lique in T 2. But then y is neighbour of 3 on P, ontrition. Conversely, let A = {C 1,...,C k } be n steroil olletion of liques, n let T be ny lique tree of G. Let T enote the subgrph of T forme by tking ll pths in T between the liques C 1,...,C k. Clerly, T is tree. Morever, we show tht C 1,...,C k re the leves of T. Otherwise, we onlue, without loss of generlity, tht C 3 belongs to the pth of T from C 1 to C 2. But then C 3 hits this pth whih ontrits Lemm 12, sine we ssume tht there is t lest one pth between C 1 n C 2 in C r (G) misse by C 3. So we let C i C i be the (unique) ege of T inient to C i, n let i be ny vertex of G in C i \C i. Further, we efine A = { 1,..., k }. Clerly, A is n inepenent set of G, sine for ll i, the lique C i is the only one in T tht ontins i. We show tht A is lso n steroil set of G. Suppose otherwise, n, without loss of generlity, ssume tht 1, 2 re in ifferent onnete omponents of G N[ 3 ]. Let P = C (1),...,C (t) be the pth in T from C (1) = C 1 to C (t) = C 2. By Lemm 12, P is misse by C 3, n hene, no ege C (i) C (i+1) of P stisfies C (i) C (i+1) C 3. Consequently, there exists x i in (C (i) C (i+1) )\C 3 for eh i {1...t 1}, n P = 1,x 1,...,x t 1, 2 is pth in G from 1 to 2. Now, rell tht 1, 2 re in ifferent onnete omponents of G N[ 3 ]. So 3 hs neighbourx i onp, n sinec 3 is lef oft, we must onluex i C 3, ontrition Goo split-minors In the previous setion, we showe tht the steroil number of G n be eue from the reue lique grph of G. To utilize this, we now investigte the struture of reue lique grphs 14